TITLE: Is it acceptable to use reduced row echelon to show basis?
QUESTION [7 upvotes]: I am asked to show that {a, b, c} forms a basis for $\Bbb R^3$. I'm just wondering if it is acceptable to use reduced row echelon to show it since it is not shown that way in the marking scheme?

$$a= \begin{pmatrix}
        2\\
        -1\\
        1 \\
        \end{pmatrix}b= \begin{pmatrix}
        1\\
        1\\
        1 \\
        \end{pmatrix}
 c= \begin{pmatrix}
        0\\
        1\\
        -1 \\
        \end{pmatrix}$$

Here's how I did it:
For the vectors to form a basis, they must be linearly independent such that
$$a_1\begin{pmatrix}
        2\\
        -1\\
        1 \\
        \end{pmatrix}+a_2\begin{pmatrix}
        1\\
        1\\
        1 \\
        \end{pmatrix}+a_3\begin{pmatrix}
        0\\
        1\\
        -1 \\
        \end{pmatrix}=\begin{pmatrix}
        0\\
        0\\
        0\\
        \end{pmatrix}$$
So, I did reduced row echelon and I got..
$$\begin{bmatrix} 2 & 1 & 0 \\-1 & 1 & 1\\  1 & 1 & -1 \\
     \end{bmatrix} \rightarrow \begin{bmatrix}
        1 & 0 & 0 \\
        0 & 1 & 0\\
        0 & 0 & 1 \\
        \end{bmatrix}$$
After that, I solve for $a_1, a_2$ and $a_3$
$$\begin{bmatrix}
        1 & 0 & 0 \\
        0 & 1 & 0\\
        0 & 0 & 1 \\
        \end{bmatrix}\cdot \begin{Bmatrix}
        a_1\\
        a_2\\
        a_3 \\
        \end{Bmatrix}=0$$
and I got $a_1=a_2=a_3=0$. Then, I conclude that {a,b,c} forms a basis.

REPLY [4 votes]: Your proof is completely correct.
As a matter of style and comprehension, I'd change two very small (purely cosmetic) things before turning it in.

When you write "So, I did reduced row echelon and I got", I'd specifically mention that you are suppressing the column of zeroes in your augmented matrix.  So I'd state "I used elementary row operations on this augmented matrix (with the column of zeroes suppressed) until getting RREF"
Once you'd gotten the RREF, I'd simply state "Converting this into a system of equations, we can see that $$\begin{cases} a_1 = 0 \\ a_2 = 0 \\ a_3 = 0\end{cases}$$  I would do this because $$\begin{bmatrix}
        1 & 0 & 0 \\
        0 & 1 & 0\\
        0 & 0 & 1 \\
        \end{bmatrix}\cdot \begin{Bmatrix}
        a_1\\
        a_2\\
        a_3 \\
        \end{Bmatrix}=0$$ is somewhat nonstandard notation.  And my way gets to the point quicker.


Edit: As hardmath pointed out to me in the comments, you should also mention at the end that because the dimension of $\Bbb R^3$ is $3$ (presumably this was proven at some point in your course?), any three linearly independent vectors will form a basis.